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Math Help - Finding the length of part of a curve.

  1. #1
    Super Member Showcase_22's Avatar
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    Finding the length of part of a curve.

    Like the title of this thread suggests, I need to find the length of a section of the curve. I've done most of the questions but I got stuck on these ones:



    I'm clearly doing the polar co-ordinate ones incorrectly, and the parametric one has me slightly confused (are the limits t=at^2 and t=0?)

    If anyone can solve any of these it would be a great help!

    If anyone wants I can post my work so far. It's just really long!

    NOTE THAT 7 HAS BEEN DONE! =D
    Last edited by Showcase_22; September 12th 2008 at 05:13 AM. Reason: typo!
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  2. #2
    Super Member Showcase_22's Avatar
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    Here's my working for the first one (ie.question 7):



    I can't really see what's wrong with my answer. It's different to the one in the book so I must be making a mistake somewhere.
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  3. #3
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    the limits for the first one should be from t=0 to t=t

    because when t=t, x=at^2 and y=2at which is what is asked
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  4. #4
    Super Member Showcase_22's Avatar
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    Ohhh!! I get it!

    Is the rest of my working fine though?
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  5. #5
    Super Member Showcase_22's Avatar
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    The book gives the answer as:



    I'm not really sure where the last part comes from. It looks a little bit like an arsinh.
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  6. #6
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    theres a problem with your substitution,

    make the substitution
     t=\sinh u

    the integral then reduces to
     2a \int_{0}^{\sinh^{-1}t} \cosh^2 u du

    calculating that gets you to the answer

    a(\sinh^{-1}t+t\sqrt{t^2+1})
    Last edited by thelostchild; September 12th 2008 at 03:55 AM.
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  7. #7
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    Just to add for the polar ones have a look here
    Pauls Online Notes : Calculus II - Arc Length with Polar Coordinates

    theres a good derivation and the example given is pretty much Q11 for you
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  8. #8
    Super Member Showcase_22's Avatar
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    did you work it all the way through?

    The second part to get the t(t^2+1)^0.5 is taking me ages!
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  9. #9
    Super Member Showcase_22's Avatar
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    I've done the integration and i've worked through it but i've hit a snag:



    Is there an easier way to work this out? I've practically covered my whiteboard in working and i'm not any closer to getting what I need to.

    My integration went like:

    Last edited by Showcase_22; September 12th 2008 at 05:01 AM.
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  10. #10
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    Quote Originally Posted by Showcase_22 View Post
    I've done the integration and i've worked through it but i've hit a snag:



    Is there an easier way to work this out? I've practically covered my whiteboard in working and i'm not any closer to getting what I need to.
    I just did it the lazy way without switching to exponentials

    use \cosh x = \sqrt{\sinh^2 x +1}

    so \cosh (\sinh^{-1}t)=\sqrt{(\sinh (\sinh^{-1}t))^2+1}=\sqrt{t^2+1}
    Last edited by thelostchild; September 12th 2008 at 05:17 AM.
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  11. #11
    Super Member Showcase_22's Avatar
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    lol! That's waaaaaay easier than doing it the way I was.

    Thanks.

    I'm working on question 11 now. I was able to get an answer for it before, but it was different to the one in the back of the book so i'll read what's at the address you posted and try again.
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  12. #12
    Super Member Showcase_22's Avatar
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    All that for the wrong answer.

    Apparently the answer is just 8a.

    ???
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